6位移法例题

位移法应用举例k11=6i例qililqΔ1图1MΔ1=13i3i①基本体系解k11Δ1+F1P=0②位移法方程③求系数、解方程k113i3iF1Pql2/8MP图ql2/8F1P=-ql2/8Δ1=-F1P/k11=ql2/48iM图ql2/163ql2/32PMMM11⑤例EI=常数①基本体系解②位移法方程k11Δ1+F1P=0③求系数、解方程k11=11ik114i3i4iΔ1=13i4i2i图1M4i2iABDECqlll△1F1P=ql2/24Δ1=-F1P/k11=-ql2/264iF1Pql2/12ql2/85/4413/1325/661/661/1322qlM图PMMM11⑤ql2/8ql2/12ql2/12图PM例EI=常数i=EI/l①基本体系解②位移法方程k11Δ1+F1P=0③求系数、解方程k11=5ik11i4iΔ1llMΔ1=1i图1M4i2iiM图PMΔ1=-F1P/k11=M/5iPMMM11⑤F1P=-MF1PM图M4M/5M/5M/52M/5只有结点集中力偶作用时，MP图=0，F1P≠0例EI=常数,i=EI/4①基本体系解②位移法方程k11Δ1+k12Δ2+F1P=0k21Δ1+k22Δ2+F2P=0③求系数、解方程k11=12ik118i4i20kN/m4m2m2m4m2EI2EIEIEI40kN10kNmΔ1=1图1M4i2i8i4ik21=4ik214iΔ2Δ1Δ2=1图2M8i6i4i2i4i）图（kNmMP26.730k12=4ik124ik22=18ik218i6i4iF2P=-3.3F2P26.730F1P=-36.7F1P26.710）图（kNmM12.42.9306.42.11.1Δ1=-3.2/iΔ2=-3.2/iPMMMM2211④结点集中力偶对MP图没有影响；对FiP有影响。①基本体系解②位移法方程k11Δ1+F1P=0lliiqΔ1Δ1=13i/l3i/l图1Mql2/8图PM③求系数、解方程F1P3ql/8F1P=-3ql/8k113i/l23i/l2k11=6i/l2例（有侧移的结构）Δ1=-ql3/16iPMMMM2211④※由弯矩图求剪力图M5ql2/163ql2/16例（有侧移的结构）①基本体系解②位移法方程k11Δ1+F1P=0lliiPiiEA=∞Δ1Δ1=1图1M3i/l3i/l3i/l3i/l③求系数、解方程k113i/l23i/l2k11=12i/l23i/l23i/l2图PMPΔ1F1PF1P=-PPΔ1=Pl2/12i只有结点集中力作用时，MP图=0，F1P≠0PMMM11④图MPl/4Pl/4Pl/4Pl/4例①基本体系解②位移法方程k11Δ1+k12Δ2+F1P=0k21Δ1+k22Δ2+F2P=0③求系数、解方程4m2i2ii2m2m20kN/mΔ1Δ2图1M3iΔ1=14i4i8ik11=19ik113i8i8ik21=012i/212i/2k21图2MΔ1Δ2=112i/212i/212i/2k22=12i6i6ik12=0k1212i/212i/2k22）图（kNmMPΔ1101010Δ2F1P=0F2P=-60F1P30301010F2PΔ1=0Δ2=5/iPMMMM2211④）图（kNmM404020例各杆EI=常数i=EI/l①基本体系解②位移法方程k11Δ1+k12Δ2+F1P=0k21Δ1+k22Δ2+F2P=0③求系数、解方程Δ1Δ2k21=06i/lk216i/l2i2i图1MΔ1=1Δ24i4i3ik11=11i3i4i4ik11lllM图2MΔ2=1Δ16i/l6i/l3i/l3i/l6i/lk22=30i/l212i/l2k2112i/l23i/l23i/l2F2P=0F2PΔ2图PMΔ1MF1P=-MF1PMk12k12=06i/l6i/lΔ1=M/11iΔ2=0PMMMM2211④图M2M/114M/114M/113M/11③求系数、解方程k11=4i1+2i2k112i14i2Δ1=ql2/12(4i2+2i1)F1P=-ql2/12k11ql2/12图1MΔ1=14i22i22i1B图PMΔ1ql2/12ql2/24BA0244124012124212121201222122121221111iiiqliMqliiiqlqliMiABAC那么如果2122212221212211121212441241211212421212120qliiiqliMqlqliiiqlqliMiABAC那么如果讨论：图Mql2/8图Mql2/24ql2/12ql2/12图Mql2/8ql2/12ql2/12PPMMMMM那么如果0111？错在哪那里图Mql2/12ql2/12ql2/24ql2/24EI→∞EI→∞例：直接画弯矩图l若结点位移趋近于零，则该位移在有限刚度的杆件上引起的内力趋近于零。例①基本体系解②位移法方程k11Δ1+k12Δ2+F1P=0k21Δ1+k22Δ2+F2P=0③求系数、解方程lllPPEI→∞EI→∞iiiiΔ1Δ2图1MΔ26i/l6i/l6i/lΔ1=16i/lk11=24i/l212i/l212i/l2k11k21=-24i/l2k2112i/l212i/l2k12=-24i/l212i/l212i/l2k12k22=48i/l2k2212i/l212i/l212i/l212i/l2图2M6i/l6i/l6i/lΔ16i/lΔ2=16i/l6i/l图PMΔ1Δ2PPF1P=-PF2P=-PPP图MPl/2Pl/2PlPlΔ1=3Pl2/24iΔ2=Pl2/12iPMMMM2211④PlE1I1→∞Pl图MMAC=MBD∵竖杆的刚度=∞∴两个横梁左端(A、B)转角相等VC=VD∑MB=0VC=VD=P/4P/2l/2EI→∞VCVDBACDl/2图MPl/8Pl/8思考题:图示结构是否可把C处的转角φC也作为基本未知量计算①基本体系解②位移法方程k11Δ1+k12Δ2+F1P=0k21Δ1+k22Δ2+F2P=0③求系数、解方程k11=8ik12=k21=2ik22=4iF1P=-ql2/12F2P=ql2/12Δ1=3l2/14iΔ2=3l2/14iΔ1Δ2图1MΔ1=14i4i2i2iΔ2=14i2i图2M图PMql2/12ql2/12qllABCk216i/l6i/lk21=04i4i3ik11=11i①基本体系解③求系数、解方程②位移法方程k11Δ1+k12Δ2+F1P=0k21Δ1+k22Δ2+F2P=0Δ2lllΔ1=13i4i4i图1MΔ2Δ1lllqq例题F1P=-ql2/8F1Pql2/12ql2/12ql2/8F22F22=-qlql/2ql/2k12=06i/l6i/lk12k2212i/l2k22=30i/l212i/l23i/l23i/l2Δ2=1Δ1ql2/12ql2/12ql2/12ql2/8图PMΔ2=1Δ16i/l6i/l6i/l图2MΔ1=ql2/88iΔ2=ql3/30i29ql2/66ql2/445ql2/4426ql2/664ql2/44ql2/6ql2/6图MPMMMM2211④F1P=-ql2/12F1Pql2/12k11=4i2ik112i③求系数、解方程Δ1=ql2/48iΔ1ql2/12ql2/24图PMΔ1=12i2i图1MPMMM11④ql2/24ql2/24ql2/24ql2/24图MθllEIΔ1Δ1=14i4i2i2i图1M4i4iθθ2iθ图M③求系数、解方程k11=8iF1△=2iθΔ1=θ/4①基本体系解②位移法方程k11Δ1+F1△=0PMMM11④3.5iθθiθ0.5iθ图M例支座位移时的计算支座位移引起的内力与刚度成正比!!弹性支座时的计算例:EI=常数，i=EI/l①基本体系解③求系数、解方程k11=6i/l2+K②位移法方程k11Δ1+F1P=0EIqEIllΔ1KΔ1=13i/l图1Mk11K3i/l23i/l2ql2/8Δ1=1图PMF1P=6ql2/8F1P3ql/83ql/8PMMM11④讨论K→∞Δ1=0M=MPql2/8ql2/8ql2/2ql2/2K→0Δ1=ql4/8iΔ1=-6ql2/8（K+6i/l2）①支座位移分组例题lllΔΔ/2Δ/2Δ/2Δ/20M正对称反对称解Δ/2Δ②取半边结构，基本体系④求系数、解方程k11=8iF1△=-3iΔ/lΔ1=3Δ/8l③位移法方程k11Δ1+F1△=0Δ=14i2ii3i1M3iΔ/lΔ3iΔ/l图PMPMMM11⑤9Δ/8l3Δ/8l12Δ/8l18Δ/8l18Δ/8l图M10℃30℃20℃30℃解①基本结构②位移法方程k11Δ1+k12Δ2+F1t=0k21Δ1+k22Δ2+F2t=04i2i3iΔ1=1③求系数、解方程k11=7i3i4ik11k21=-6i/l6i/lk21Δ1Δ2Δ1Δ2=16i/l3i/l6i/lk12=k21=-6i/lk126i/lk22=k12=-15i/l212i/l2k223i/l2求自由项1）平均温度t0变化)(15)(20)(200lllltDVCHCViliMiliMMDVCVtCDCHtCAtAC15/)(3120/600012012015iMt图020℃15℃20℃ABCDΔCHΔCVΔDV平均温度t0变化2）温度变化ΔtihtEIMihtEIMihtEIMMtBDtCDtCAtAC150233002320020℃10℃20℃ABCD温度变化之差Δt3020iMt10图1512012015iMt图03020iMt10图15liFiFtt9020521l232002384521tMMMM2211④i236870i232850i234020